Erdős Problem #948

Is there a function \(f(n)\) and a \(k\) such that in any \(k\)-colouring of the integers there exists a sequence \(a_1 < a_2 < \cdots\) such that \(a_n < f(n)\) for infinitely many \(n\) and the set \[\left\{ \sum_{i \in S} a_i : \textrm{finite nonempty } S \right\}\] does not contain all colours? A question of Erdős [Er77c] and Erdős and Galvin [ErGa91]. The answer is no: GPT-5.5 Pro (prompted by Price) showed that for every \(f\) there is a colouring of the integers such that the finite sums of every such sequence use all colours. The linked formal proof (Codex and GPT-5.6 Sol) gives, for every \(f\) and \(k \ge 1\), a colouring \(c : \mathbb{Z} \to \{0, \ldots, k - 1\}\) such that for every strictly increasing \(a\) with \(a_n < f(n)\) infinitely often and every colour, some nonempty finite sum of the \(a_i\) has that colour.